Is Quantum Teleportation possible without a classical channel?

(This was first posted to a newsgroup but there were no replies;
so I am trying to post it here).

I understand quantum teleportation works on qubits, but i can also
work on regular bits. Suppose Alice has a message bit S. S is
repeatable, because it is a regular bit. Now Alice wants to send it to
Bob and she is willing to repeat the message 1,000,000 times.

So, she teleports it to Bob. Both Alice and Bob receive, from a
middleman Moe, a stream of entangled particles. In total Moe sends
1,000,000 bursts of entangled photons--each burst consists of at least
one entangled photon going left to Alice, and one going right to Bob.

Quantum Teleportation was designed to work on qubits, which in general
cannot be cloned. But, suppose we have a classical bit S we want to
send from Alice to Bob. Then we can try to send it 1,000,000 times.
For each attempt, there is an entangled particle distributed from Moe
to Alice and to Bob.

So Alice manipulates its entangled qubit with its message bit S, and
the resulting qubit is interrogated to produce two classical bits X
and Y. Normally, to "teleport" information from Alice to Bob, Bob
receives an entangled particle from Moe (the middleman) and then
detects the entangled qubit two different ways, depending on the
classical information X and Y that was sent.

My question is: what if we don't send any classical bits? There are 2
bits of classical information per burst (attempted message send). Thus
there is a 25% chance we will receive the message that was sent!

If we send the message 1,000,000 times, then won't the message be
received 250,000 times? That's pretty good.

My question is, what happens the other 750,000 times?

Basically, my setup (for review) is a simple quantum teleportation
setup, except instead of using the classical bits sent by Alic to Bob,
we generate two random numbers (bits) at Bob and use them instead!

My hope - unfounded at present - is that 750,000 times when the
classical bits guessed are wrong, we receive a random BIT on the
output. I might be wrong. Maybe instead we receive a random QUBIT.

But if we have "signal" 250,000 times and "noise" 750,000 times, and
half the noise is 0 and half the noise is 1, that means we receive
250,000 + 750,000 / 2 = 250,000 + 375,000 = 625,000 signals and
375,000 noise bits.

Thus. If I send a message from Alice to Bob 1,000,000 times, I expect
to receive the message over 60% of the time; there will be the wrong
message (opposite bit) under 40% of the time. Can't a computer easily
tell whether 0 occurred more frequently than 1 at Bob's receiving
station, or whether 1 occurred more frequently than 0?

But this cannot be right. If it were, strange things indeed would be
possible. Quantum teleportation was not designed to be used in this
way. But what if it worked? Then maybe, just maybe (I'm speculating
here), we can distribute entangled particles across time -- the future
can receive one entangled particle and the past, the other. This can
be done by slowing down light, or sending it through 10km of circular
fiber optics channel (John Cramer's idea).

Normally a classical channel is needed to send a message, but since
we're sending a classical bit we can send it 1,000,000 times and let
majority rule. Without needing a classical channel to send information
from the future to the past, we can rely on entanglement.

And that is where we get ourselves into trouble. For, what if it
worked? If classical bits can be sent from the future to the past,
what does it mean? My question is as follows. If I detect a random
qubit, maybe there are now two universes. Thus the past had one
universe, there are two futures. If the future sends a message to the
past, which future gets to talk?

Quantum computers have severe limitations. For example, if this
universe splitting were real, the output bit (computed using
reversable logic using quantum gates) exists in two universes. We can
CNOT them together, i.e. I can tell whether or not f(0) = f(1). That's
all fine, but I want to perform nonlinear operations -- rather than
xor'ing two bits output from a quantum circuit, I want to OR them
together (this will make it possible to solve exponential-time
problems in polynomial time, i.e. subset sum, encryption cracking, and
so on).

But if I could split this universe into two universes and then send
messages from the future to the past, what if I could receive two
messages through two separate channels? The past receives two messages
then! The idea is to OR them together and then send them to the past's
past. By repeating this process, I can find any f(X) where X is a
vector of unknown bits, and f is some boolean function that returns 1
if X is in the set and 0 if X is not in the set.

Before I try to figure out how to build an "exotic network" (i.e.
quantum computer that sends messages from the many futures to a single
past to exploit massive parallelism), I want to make sure my idea is
grounded in reality.

Anyone with a better understanding of quantum mechanics out there,
care to speculate about what the outcome of my proposed experiment
would be?

All we need to do is prepare a message to teleport that is exactly 0,
then teleport it from Alice to Bob with Bob not really using the
classical bits Alice is sending, but instead using random bits for the
two classical bits sent over the classical channel. We repeat the
experiment 1 million times. Then we change the message from being
exactly 0 to being exactly 1 and do the experiment another 1 million
times.

If anyone out there knows of any labs with access to quantum
teleporation equipment, what do you think ? It doesn't sound like it
would be expensive. Since the classical channel is essentially random,
I was wondering if maybe we could make it cheaper, by always using 0
for the classical bits.

(This was first posted to a newsgroup but there were no replies; so I am trying to post it here).

Welcome to PF.

But if we have "signal" 250,000 times and "noise" 750,000 times, and half the noise is 0 and half the noise is 1, that means we receive 250,000 + 750,000 / 2 = 250,000 + 375,000 = 625,000 signals and 375,000 noise bits.

I'm afraid you've managed to confuse yourself here when switching from two-bit probabilities to single-bit probabilities.
Whichever two-bit signal I choose (00,01,10,11) would get 'sent' 1/4 of the time, so far so good.

But how would the single bits be distributed? If I pick 00 as my 'signal', then the patterns 10, 01, and 11 are 'noise'. So 4/6 of the 'noise' would be ones, and (4/6)x(3/4) = 1/2 of the total signal would be ones. If I pick 11 as my 'signal', the noise is only 2/6 ones, which works out to 1/4 the total being ones from the noise, and since the signal 1/4 is all ones in this case, it works out to a 50%-50% total again. And if I choose 01 or 10, half the noise is ones, as well as half the signal.

So half the stuff being received is 1s and half 0s, and 1/4 is each of 00,01,10,11. Your message gets received 1/4 of the time. But you have no way of knowing which 1/4 it is.

1. Generate entangled particles, send half of them to Alice who has the message S to send and half to Bob who wants to receive the message.
2. Let Alice prepare a qubit Q such that it represents the classical bit S with 100% probability
3. Let Alice operate on the qubit Q and the entangled particle.
4. Let Alice detect the result of this operation two different ways, yielding 2 classical bits.
5. Let Bob receive the two classical bits. This is the classical channel.
6. Let Bob also receive the entangled particle. This is the quantum channel.
7. Let Bob detect the entangled particle two different ways depending on the two classical bits, and put the information together.
8. Bob has now received S.

As you can see, there are two separate channels here--the quantum and the classical channel. The classical channel is essentially "random information" and does not reflect S.

For example, if I want to send S=1 repeatedly, the classical bits might be 01, 11, 10, 10, 00, 01, ... they are seemingly random. The number of times 1 appears in the classical stream has nothing to do with the message that was sent.

The question I have is what message will Bob receive in step 8 if instead of using the 2 classical bits transmitted to him in step 5, we use two random classical bits for the classical channel?

alxm, you seem to be forgetting about the quantum channel. I agree that if I have a sender, message, and receiver and I somehow encode my 1 bit that I want to send, in 2 bits, and the receiver generates its 1 bit output not from the 2 bits it received but from 2 random bits, then the output will be garbage.

But what about the entangled particles we have here? That changes things. The idea was that if we repeat the experiment a lot of times, the message seems to be received more than 60% of the time. The opposite of the message is received under 40% of the time.

Here is my reasoning -- but don't forget, I attempt to use quantum entanglement as a second channel!

Given 1,000,000 experiments, 1/4 of the time the two classical bits generated randomly at Bob's end will match the two classical bits detected at Alice's end. Thus quantum entanglement works as expected and Bob receives Alice's message S. We thus know Bob receives Alice's message at least 25% of the time.

What happens the other 75% of the time? Since I don't know, I made an assumption. If you know better, please explain.

My assumption is, 75% of the time you get either a 1 or 0 with equal probability.

That is, if we try to send S, we receive S with probability 0.25 + 0.75 / 2 and we receive S' (the opposite bit for S) 0.75 /2 of the time.

But since 0.25 + 0.75 / 2 = 0.625 and 0.75 / 2 = 0.375 it looks like we receive the single bit S we tried to send, over 60% of the time, in theory!

Since we are trying to send a classical bit, it's repeatable and we can try the experiment 1,000,000 times in parallel, then let majority rule to receive the message.

If this doesn't work, why not? One guess is that the entangled particle contains information such that if you guess the two classical bits right (we do, 1/4 of the time), then out comes the right answer. But if we guess wrong, it might still contain information about the message, in a way so that the remaining 3/4 of the time, instead of it being 50% S and 50% S', it's 1/3 S and 2/3 S'. Since 3/4 * 1/3 = 1/4, and 1/4 + 1/4 = 1/2, it follows that if this is what quantum teleportation does, then by guessing the two classical bits we get S half the time and S' (the opposite of S) the other half the time.

The above seems strange to me. Are the laws of physics such that if I teleport a qubit which is either 1 with 100% probability or 0 with 100% probability, and the classical channel is not right (a 3/4 chance if we guess the classical channel, which has 2 bits), then how does the entangled particle, when detected in the way we choose to detect it, know it's supposed to be S' 2/3 of the time and S 1/3 of the time, so as to make our signal worthless?